An arrow is shot at an angle 31 ◦ with the
horizontal. It has a velocity of 47 m/s.
How high will the arrow go? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m
To find the height the arrow will reach, we need to use the equations of motion for projectile motion.
First, let's break down the initial velocity into its horizontal and vertical components. We can use trigonometry to do this. The horizontal component is given by:
Vx = V * cos(theta)
where V is the magnitude of the velocity (47 m/s) and theta is the angle (31 degrees).
Vx = 47 m/s * cos(31 degrees)
Vx ≈ 40.36 m/s
Similarly, the vertical component is given by:
Vy = V * sin(theta)
Vy = 47 m/s * sin(31 degrees)
Vy ≈ 24.22 m/s
Since we want to find the height the arrow will reach, we need to find the time it takes for the arrow to reach its peak height. We can use the formula:
Vy = Voy + a * t
where Voy is the initial vertical velocity (Vy), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
0 m/s = 24.22 m/s + (-9.8 m/s^2) * t
Simplifying the equation gives us:
-24.22 m/s = -9.8 m/s^2 * t
Solving for t gives us:
t ≈ 2.47 seconds
Now that we know the time it takes for the arrow to reach its peak height, we can find the height by using the vertical position formula:
y = Voy * t + (1/2) * a * t^2
y = 24.22 m/s * 2.47 s + (1/2) * (-9.8 m/s^2) * (2.47 s)^2
y ≈ 29.96 meters
Therefore, the arrow will reach a height of approximately 29.96 meters.
the initial vertical speed is 47 sin 31° = 24.2 m/s
So, the height
y = 24.2t - 4.9t^2
Now use your algebra to find the vertex of that parabola. That will be the maximum height.